(101)

Equations (100), (101) indicate that the amplitude of the time-averaged energy equals one half of the amplitude of the energy function given by Eq. (98) and is proportional to the frequency squared.

To obtain the *power*, substitute Eq. (97) into Eq. (83) and get

$P(x,t)=ce(x,t)=mc{\omega}^{2}{\stackrel{\u02c6}{u}}^{2}{\mathrm{sin}}^{2}(\gamma x-\omega t)$ (102)

(102)

The power amplitude is

$\stackrel{\u02c6}{P}=mc{\omega}^{2}{\stackrel{\u02c6}{u}}^{2}$ (103)

(103)

To calculate the time-averaged power, recall Eq. (81), i.e.,

$P={\int}_{A}\overrightarrow{t}\cdot \dot{\overrightarrow{u}}dA=-\sigma \dot{u}A=$

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